Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(X,d)$ be a metric space and $ a \in X $

Show that $f(x) = d(x,a)$ is a Lipschitz function from X to $\Re$

Use this fact to show $S$ is an open subset of $X$:

Let $(X,d)$ be a metric space and $a \in X, r > 0.$ Let $S$={$x \in X : d(a,x) > r$}

>Lipschitz Defn:

Let $f$ be a function from $(X,d)$ to $(Y,\rho)$

$f$ is said to be Lipschitz if:

$\exists$ K $\geq$ 0 s.t. $\rho( f(x_1) , f(x_2) ) \leq$ Kd($x_1$,$x_2$) $\forall$ $x_1 x_2 \in X$

>Continuity Defn:

Let $(X,d)$ and $(Y,\rho)$ be metric spaces, and $f: X \to Y$

Then $f$ is continuous $iff$ f$^{-1}$($B$) = { $x \in X : f(x) \in B $} is an open subset of $X$, whenever $B$ is an open subset of $Y$

>Triangle Inequality for a Metric:

$d(x,z) \leq d(x,y) + d(y,z)$

My first question is, am I thinking about this correctly?

If $f(x)$ is a Lipschitz function, it is uniformly continuous

If $f(x)$ is uniformly continuous, it is continuous at every point

If $f(x)$ is continuous at every point $\Rightarrow$ the inverse images of open sets are open

Hence $S$ is an open subset of $X$.

Secondly, how would you formally construct this?

We know $f(x) = d(x,a)$ s.t. $f:X \to \Re$

By definition of Lipschitz we have $\exists K \geq 0$ s.t. $\rho(f(x_1),f(x_2)) \leq Kd(x_1,x_2) \forall x_1,x_2 \in X$

& since our codomain is $\Re$

$\rho(f(x_1),f(x_2)) = | d(x_1,a) - d(x_2,a) | \leq Kd(x_1,x_2)$

Taking $K=1$


Help in finishing this off would be appreciated.


share|cite|improve this question
Why did you use LaTeX only for part of the formatting? Please typeset all the math with LaTeX. – TMM Oct 31 '12 at 21:55
up vote 1 down vote accepted

You can take $K=1$ in the definition.

For all $x_1,x_2\in X$, we have: $$d(x_1,a)\leq d(x_1,x_2)+d(x_2,a)$$ and $$d(x_2,a)\leq d(x_1,x_2)+d(x_1,a)$$ by the triangle inequality. We may rewrite these as $$d(x_1,a)-d(x_2,a)\leq d(x_1,x_2)$$ and $$d(x_2,a)-d(x_1,a)\leq d(x_1,x_2).$$ These two inequalities together mean precisely that $$|d(x_1,a)-d(x_2,a)|\leq d(x_1,x_2)$$ which is Lipschitz continuity of $f$ with constant $K=1$.

For the first question, you are correct. $S$ is open because $S=f^{-1}(r,\infty)$ and $f$ is continuous.

share|cite|improve this answer
I'm having trouble following this. I made a typo in the Lipschitz Defn above that I have corrected and I think you followed through with it. – user42538 Oct 31 '12 at 22:27
@user42538: I have reworded it a bit. I hope it's easier to follow now. In case something is not clear, you are welcome to ask. – Dejan Govc Oct 31 '12 at 22:36
Ah, completely my bad. Although I did make a typo, for some reason I was considering a as the other point, rather than looking at it as f(x_1) and f(x_2). Thanks. – user42538 Nov 1 '12 at 0:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.